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# Finding the Solution to the Initial Value Problem 7

DIFFEQ-LY0PM1

The solution to the initial value problem:

$$X'=\left(\begin{array}{cc} 1 & 1 \\\ 4 & -2 \end{array}\right) X, \ X(0)=\left(\begin{array}{c}1 \\\ -2\end{array}\right)$$

...is:

A

$X=\cfrac{2}{5}\left(\begin{array}{c} -1 \\\ 1\end{array}\right)e^{2t}+\cfrac{3}{5}\left(\begin{array}{c} 1 \\\ -4 \end{array}\right)e^{-3t}$

B

$X=\cfrac{2}{5}\left(\begin{array}{c} 1 \\\ -1\end{array}\right)e^{2t}+\cfrac{3}{5}\left(\begin{array}{c} 1 \\\ -4 \end{array}\right)e^{-3t}$

C

$X=\cfrac{2}{3}\left(\begin{array}{c} 1 \\\ 1\end{array}\right)e^{2t}+\cfrac{3}{5}\left(\begin{array}{c} 1 \\\ 4 \end{array}\right)e^{-3t}$

D

$X=\cfrac{2}{3}\left(\begin{array}{c} 1 \\\ 1\end{array}\right)e^{3t}+\cfrac{3}{5}\left(\begin{array}{c} 1 \\\ -4 \end{array}\right)e^{-2t}$

E

None of the above